JEE Physics Wave Optics MCQs Set 09

Answer: (b) 121 : 81

Explanation:
1. Amplitude ratio \( =\sqrt{100}:\sqrt1=10:1 \).
2. \( \frac{I_{max}}{I_{min}}=\frac{(10+1)^2}{(10-1)^2}=\frac{121}{81} \).

Teacher's Note:
1. Intensity is proportional to the square of amplitude.
2. Imax/Imin = (a1 + a2)²/(a1 − a2)².

Answer: (b) \( \frac{2\pi}{3} \)

Explanation:
1. \( y_2=A_2\cos\left(3\omega t+\frac\pi6\right)=A_2\sin\left(3\omega t+\frac\pi6+\frac\pi2\right) \).
2. Phase difference \( =\frac\pi6+\frac\pi2=\frac{2\pi}{3} \).

Teacher's Note:
1. Convert cosine to sine by adding π/2 to the argument.
2. The phase difference is the difference between the two arguments.

Answer: (b) 180°

Explanation:
1. Phase difference \( =\frac{2\pi}{\lambda}\times\frac\lambda2=\pi \) rad = 180°.

Teacher's Note:
1. A path difference of one wavelength corresponds to 360°.
2. So half a wavelength corresponds to 180°.

Answer: (b) the separation of the slits

Explanation:
1. \( \beta=\frac{\lambda D}{d} \).
2. Fringe width increases with λ and D and decreases with the slit separation d.
3. So decreasing d increases β.

Teacher's Note:
1. β is inversely proportional to slit separation.
2. Changing the width of each slit alters the brightness and the diffraction envelope, but not the fringe spacing.

Answer: (d) by using white light instead of single-wavelength light

Explanation:
1. With white light every colour forms its own fringes, but all colours have a bright maximum at the centre (zero path difference).
2. So the central fringe is white while the others are coloured, which makes it identifiable.

Teacher's Note:
1. With monochromatic light the fringes are almost equal in width and brightness, so the centre cannot be recognised.
2. With white light the order-0 fringe is the only colourless one.

Answer: (b) 7

Explanation:
1. Maxima need \( d\sin\theta=n\lambda \), so \( n\le\frac d\lambda=\frac{700}{200}=3.5 \).
2. Possible orders: \( n=0,\pm1,\pm2,\pm3 \).
3. Total number of maxima \( =2\times3+1=7 \).

Teacher's Note:
1. The largest possible order is the integer part of d/λ.
2. Count both sides of the central maximum.

Answer: (b) 2

Explanation:
1. Equal positions: \( n\times7800=(n+1)\times5200 \).
2. \( 2600n=5200\Rightarrow n=2 \).

Teacher's Note:
1. The n-th bright fringe is at nλD/d.
2. The ratio 7800 : 5200 is 3 : 2, so the 2nd red and 3rd blue coincide.

Answer: (c) the fringe next to the central will be red

Explanation:
1. Fringe positions are proportional to λ, so the first-order fringe of violet (shortest λ) lies closest to the centre.
2. So the fringe next to the central is violet, and red lies farthest out in each order.
3. The statement that the fringe next to the central is red is false.

Teacher's Note:
1. The central maximum is white because all colours have zero path difference there.
2. In each order, the colours run from violet (nearest the centre) to red (farthest).

Answer: (c) I/4

Explanation:
1. With both slits: amplitude \( 2a \), intensity \( I\propto4a^2 \).
2. With one slit: amplitude \( a \), intensity \( \propto a^2=\frac I4 \).

Teacher's Note:
1. Intensity is proportional to the square of the amplitude, and the amplitudes add at the central maximum.
2. Covering one slit removes the interference, leaving a quarter of the central intensity.

Answer: (c) \( 1.25\times10^{-4}\ \text{m} \)

Explanation:
1. Half intensity: \( \cos^2\frac\phi2=\frac12\Rightarrow\phi=\frac\pi2 \), i.e. path difference \( \frac\lambda4 \).
2. \( \frac{yd}{D}=\frac\lambda4\Rightarrow y=\frac{\lambda D}{4d}=\frac{5\times10^{-7}\times1}{4\times10^{-3}}=1.25\times10^{-4}\ \text{m} \).

Teacher's Note:
1. Half the maximum intensity corresponds to a phase of π/2, a quarter of a wavelength in path.
2. y is a quarter of the fringe width β = λD/d = 0.5 mm.

Answer: (c) \( \lambda/2 \)

Explanation:
1. Half intensity needs a path difference of \( \frac\lambda4 \) (phase \( \frac\pi2 \)).
2. The sheet introduces extra path \( (\mu-1)t=0.5t \).
3. \( 0.5t=\frac\lambda4\Rightarrow t=\frac\lambda2 \).

Teacher's Note:
1. A transparent sheet of thickness t adds an extra optical path of (μ − 1)t.
2. Set this equal to λ/4 for half intensity.

Answer: (c) \( 3I_0/4 \)

Explanation:
1. Path difference \( \Delta=\frac{xd}{D}=\frac{4\times10^{-5}\times2.5\times10^{-3}}{1}=10^{-7}\ \text{m}=\frac\lambda6 \).
2. Phase difference \( \phi=\frac{2\pi}\lambda\cdot\frac\lambda6=\frac\pi3 \).
3. \( I=I_0\cos^2\frac\phi2=I_0\cos^230°=\frac{3I_0}{4} \).

Teacher's Note:
1. Express the path difference as a fraction of λ first.
2. Intensity at phase φ is I0 cos²(φ/2).

Answer: (c) 599

Explanation:
1. Maxima satisfy \( d\sin\theta=n\lambda \). At the limit \( \theta=30° \): \( n=\frac{0.3\times10^{-3}\times0.5}{500\times10^{-9}}=300 \).
2. The order 300 occurs exactly at 30°, which is excluded, so \( |n|\le299 \).
3. Number of maxima \( =2\times299+1=599 \).

Teacher's Note:
1. The maximum order allowed by the angular limit is d sin 30°/λ = 300, but the endpoint itself is excluded.
2. Count orders on both sides plus the central one.

Answer: (b) 8

Explanation:
1. Extra path difference introduced \( =(1.7-1.4)t=0.3t \).
2. The central fringe moves by 5 fringes, so \( 0.3t=5\lambda=5\times4.8\times10^{-7}=2.4\times10^{-6}\ \text{m} \).
3. \( t=8\times10^{-6}\ \text{m}=8\ \mu\text{m} \).

Teacher's Note:
1. A shift of the central fringe by n fringes means an extra optical path of nλ.
2. Net extra path is (μ2 − μ1)t because both slits carry a film of the same thickness.

Answer: (d) be four times

Explanation:
1. \( \beta=\frac{\lambda D}{d} \).
2. With \( d\to\frac d2 \) and \( D\to2D \): \( \beta'=\frac{\lambda(2D)}{d/2}=4\beta \).

Teacher's Note:
1. Fringe width is directly proportional to D and inversely proportional to d.
2. Both changes increase β, so the factors multiply: 2 × 2 = 4.

Answer: (a) 1 mm

Explanation:
1. Adjacent bright fringes are separated by one fringe width.
2. The nearest bright fringe to the central one is the first order, at \( \beta=1\ \text{mm} \).

Teacher's Note:
1. Distance between consecutive bright fringes equals β.
2. Half a fringe width (0.5 mm) would locate the first dark fringe instead.

Answer: (a) increase

Explanation:
1. \( \beta=\frac{\lambda(a+b)}{d} \), which grows as the screen distance b increases.
2. So the fringes become wider.

Teacher's Note:
1. Fringe width is proportional to the distance between the sources and the screen.
2. The separation d of the virtual sources does not depend on the screen position.

Answer: (a) fringes will be closer

Explanation:
1. Deviation \( \delta=(\mu-1)\alpha \) increases with the prism angle, so the virtual sources are farther apart: \( d=2a\delta \).
2. \( \beta=\frac{\lambda D}{d} \) decreases, so fringes get closer.

Teacher's Note:
1. A larger angle bends the rays more and separates the two virtual sources more.
2. Larger source separation means narrower fringes.

Answer: (a) 4780 Å

Explanation:
1. Fringe widths: \( \beta_1=\frac{23}{20}=1.15\ \text{mm} \) and \( \beta_2=\frac{28}{30}=0.933\ \text{mm} \).
2. Since \( \beta\propto\lambda \): \( \lambda_2=5890\times\frac{0.933}{1.15}\approx4780\ \text{Å} \).

Teacher's Note:
1. With the same set-up, β is proportional to λ.
2. Find each fringe width from the distance and number of fringes.

Answer: (a) 225 nm

Explanation:
1. Reflection at air-film (1.00 to 1.33) and at film-glass (1.33 to 1.50) both occur from a denser medium, so both rays suffer a phase change of π and the net phase shift from reflection is zero.
2. Strong reflection needs \( 2\mu t=m\lambda \), and the least thickness has \( m=1 \).
3. \( t=\frac{\lambda}{2\mu}=\frac{600}{2\times1.33}\approx225\ \text{nm} \).

Teacher's Note:
1. When both reflections involve a phase change (or neither does), constructive interference needs 2μt = mλ.
2. With an anti-reflection coating the condition would be 2μt = (m + 1/2)λ for minimum reflection.